The lifespans of seals in a particular zoo are normally distributed. The average seal lives $12.2$ years; the standard deviation is $2.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a seal living less than $7.4$ years.
Solution: $12.2$ $9.8$ $14.6$ $7.4$ $17$ $5$ $19.4$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $12.2$ years. We know the standard deviation is $2.4$ years, so one standard deviation below the mean is $9.8$ years and one standard deviation above the mean is $14.6$ years. Two standard deviations below the mean is $7.4$ years and two standard deviations above the mean is $17$ years. Three standard deviations below the mean is $5$ years and three standard deviations above the mean is $19.4$ years. We are interested in the probability of a seal living less than $7.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the seals will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the seals will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $7.4$ years and the other half $({2.5\%})$ will live longer than $17$ years. The probability of a particular seal living less than $7.4$ years is ${2.5\%}$.